Question

Rozwiąż równanie:
3(x+2)²+(2x-1)²-7(x+3)(x-3) = 28
x-3/2+x = 4x-1/2-0,25(2x+3)
x-3/2+x=4x-1/2-0,25(2x+3)

Answer 1

3(x+2)²+(2x-1)²-7(x+3)(x-3) = 28
3(x²+4x+4)+(4x²-4x+1)-7(x²-3²)=28
3x²+12x+12+4x²-4x+1-7x²+63=28
8x=-48 /:8
x=-6

x-3/2+x = 4x-1/2-0,25(2x+3)
(x-3)*[2-0,25(2x+3)]=(2+x)(4x-1)
(x-3)(2-0,5x-0,75)=8x-2+4x²-x
(x-3)(1,25-0,5x)=4x²+7x-2
1,25x-0,5x²-3,75+1,5x=4x²+7x-2
-4,5x²-4,25x-1,75 =0 i dalej trzeba liczyc delte, x₁ i x₂

Answer 2

3(x+2)²+(2x-1)²-7(x+3)(x-3) = 28
3(x²+2x+4)+4x²-x+1-7(x²-3x+3x-9)=28
3x²+6x+12+4x²-x+1-7x²+21x-21x+63=28
5x=-48
x=-9,6

x-3/2+x = 4x-1/2-0,25(2x+3)
2x-3/2=4x-1/2-0.5-0.75
2x-4x+0,5x=-1/2-1,25+3/2
-1,5x=1/4
x=-0,16


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